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<title>Simulations for Statistical and Thermal Physics</title>

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<h3 style="text-align:center;">Ideal Bose gas</h3>

<p class="header_title">Introduction</p>

<p>The motivation for discussing the ideal Bose
gas is that it exhibits
Bose-Einstein condensation. The original prediction of
Bose-Einstein condensation by Satyendra Nath Bose and Albert
Einstein in 1924 was considered by some to be a mathematical
artifact. In the 1930s Fritz London realized
that superfluid liquid helium could be understood in terms of
Bose-Einstein condensation. However, the analysis of superfluid
liquid helium is complicated by the fact
that the helium atoms in a liquid strongly interact with one
another. In 1995 several
groups used laser and magnetic traps to create a
Bose-Einstein condensate of alkali atoms at approximately
10<sup>-6</sup>K. In these systems the interaction between the
atoms is very weak so that the ideal Bose gas is a good
approximation and is no longer only a textbook
example.</p>

<p>&nbsp;&nbsp;&nbsp;&nbsp;The behavior of an ideal Bose gas can be understood by calculating
N(T,V,&#956;), the mean number of particles for given values of the temperature T, volume V, and chemical potential &#956;. In the grand canonical ensemble N is given by</p>
<p class="center">
<img src="nbarbose.jpg" alt="" align="middle" >,
</p><p>
where n(&#949;) is the mean number of particles with energy &#949; and g(&#949;) is the density of states. For a system of particles in three dimensions we know that g(&#949;) &#8733; &#949;<sup>1/2</sup>. We substitute the form of g(&#949;) for bosons and the form of n(&#949;) and write N(T,V,&#956;) as</p>
<p class="center">
<img src="nbarbose2.jpg" alt="" align="middle" >,
</p><p>
where D = 2(2m)<sup>3/2</sup>&#960;/h<sup>3</sup>.</p>

<p>&nbsp;&nbsp;&nbsp;&nbsp;To
understand the nature of an ideal Bose gas at low temperatures,
suppose that the mean density of the system is fixed and consider the
effect of lowering the temperature. The
correct choice of &#956; gives the desired value of
&#961; when substituted into</p>
<p class="center">
<img src="rhobose.jpg" alt="" align="middle" >.
</p>
<p>Let us define &#961;* = &#961;/D so that</p>
<p class="center">
<img src="rhobose2.jpg" alt="" align="middle" >.
</p>
<p>
We will study the behavior of &#956; as a function of the temperature T for a particular value of &#961;*.</p>
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<p class="header_title">Problems</p>

<ol>
<li>In the high temperature limit, the chemical potential &#956; must be given by its classical value, which is negative and large in magnitude. Investigate numerically how &#956; changes as you decrease the temperature. The program evaluates the integral for &#961;* for a given value of &#946; and &#956;. The goal is to find the value of &#956; for a given value of T that yields the desired value of &#961;*.

<br>&nbsp; &nbsp;&nbsp;&nbsp;Choose  &#961;* = 1 and  T = 10 and first choose &#956; = -10. What is the computed value of the integral? Do you have to increase or decrease the value of &#956; to make the computed value of the integral closer to &#961;* = 1? By trial and error, you should find that &#956; &#8776 -33.4. Next choose T = 5 and find the value of &#956; needed to keep &#961;* fixed at &#961;* = 1. Does &#956; increase or decrease in magnitude? You can generate a plot of &#956; versus T by clicking on the <tt>Accept parameters</tt> button.</br></li>

<li>As T is decreased at constant density, does the magnitude of &#956; increase or decrease? What is the implication of this dependence as T is lowered further?</li>
</ol>

<p class="header_title">References</p>

<ul>

<li>The 2001 Nobel Prize for Physics was awarded to
Eric Cornell, <a href="http://cua.mit.edu/ketterle_group/">Wolfgang Ketterle</a>, and <a href="http://www.colorado.edu/physics/2000/bec/">Carl Wieman</a> for achieving
Bose-Einstein condensation in dilute gases of alkali atoms and for
early fundamental studies of the properties of the condensate. Another excellent resource is at <a href="http://bec.nist.gov/">NIST</a>.</li>

</ul>

<p class = "small">Updated 2 May 2007.</p>
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